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A conveyor belt is moving at a constant speed of 2m/s. A box is gently dropped on it. The coefficient of friction between them is µ = 0.5. The distance that the box will move relative to belt before coming to rest on it taking g = 10 ms–2, is
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- 1.2 m
- 0.6 m
- zero
- 0.4 m
Correct Option: D
Frictional force on the box f = µmg
∴ Acceleration in the box
a = µg = 5 ms–2
v² = u² + 2as
⇒ 0 = 2² + 2 × (5) s
⇒ s = – 2/5 w.r.t. belt
⇒ distance = 0.4 m
